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2 years ago

It takes Santa a total distance of 1800 miles to cover a portion of his

Mathematics

2 answers:

alexandr402 [8]2 years ago

8 0

Answer:

the answer is 600mph

Step-by-step explanation:

because 1800miles/3 hours=600mph

olya-2409 [2.1K]2 years ago

4 0

Answer:

600 mph

Step-by-step explanation:

In order to find the answer you must solve using this formula:

S=D/T

Where S is speed, D is distance, and T is time

You have time and distance, so you have to plug those into the formula:

S=1800/3

Then all you have to do is divide 1800/3 and get 600 for speed.

You might be interested in

Find the sum of 12a + 7b, −6a − 9, and 14a − 12b.

Ivan

The answer is: [A]: " 20a − 5b − 9 " .
______________________________________
Explanation:
_______________________________________________
(12a <span>+ 7b) + (−6a − 9) + (14a − 12b) =

12a </span><span>+ 7b + 1(−6a − 9) + 1(14a − 12b) =
</span>
12a + 7b + (1*-6a) + (1*-9) + (1*14a) + (1* -12b) =

12a + 7b − 6a − 9 + 14a − 12b = ?

Combine the "like terms:

12a − 6a + 14a = 20a ;

7b − 12b = - 5b ;

and then we have "-9" ;
_____________________________________________________
So, write as: " 20a − 5b − 9 " ; which is: Answer choice: [A].
______________________________________________________

8 0

3 years ago

Read 2 more answers

Ky buys three tshirts originally priced at 49.96the stores discount applys 7.1 tax

Olin [163]

Answer:

Step-by-step explanation:

49-7=42

7 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .